Navigation Problems: Relative Velocity, PNR and CP

Once closing or opening speed is known, the rest is an ordinary speed, distance and time calculation. Keep the geometry separate from the arithmetic.

Closing speed

Two aircraft are 460 NM apart and fly directly towards each other at groundspeeds of 250 kt and 210 kt. Closing speed is 460 kt, so they meet after 460 / 460 = 1 hour.

Meeting point

In the preceding example, the first aircraft flies 250 NM and the second flies 210 NM. The two distances add to the original 460 NM separation.

Overtaking

A faster aircraft at 220 kt is 80 NM behind an aircraft at 180 kt on the same track. Relative closing speed is 40 kt. Catch-up time is 80 / 40 = 2 hours. The faster aircraft covers 440 NM before overtaking; the slower covers 360 NM from its own initial position.

Opening motion

If the faster aircraft is already ahead, the same speed difference becomes opening speed. After 30 minutes at an opening rate of 40 kt, separation increases by 20 NM.

When tracks are not collinear, draw the geometry. Relative bearings, elapsed distance and the internal angles of the triangle often reveal closest approach without a long calculation.

Triangular Problems

Choose a convenient north reference, draw successive relative-bearing lines and join the aircraft positions by the distance travelled. The closest point of approach is the perpendicular distance from the beacon or other traffic to the relative path.

Worked closest-point example

A beacon is at relative bearing 270. Three minutes later it is at relative bearing 225 while groundspeed is 180 kt and heading is constant. The aircraft covers 180 x 3 / 60 = 9 NM. The geometry contains two 45-degree angles, so the closest-approach distance is also 9 NM.

Two radial and range fixes

An aircraft is on the 310 radial at 10 NM, then ten minutes later on the 040 radial at 10 NM. The radii form a right angle and an isosceles triangle. Distance between fixes is the square root of 10 squared plus 10 squared, or 14.14 NM. The mean track is 085 M and groundspeed is about 85 kt.

Closest approach from relative velocity

In a full vector problem, extend the relative track from the initial relative position. Drop a perpendicular from the origin to that line. The perpendicular is closest-approach distance, and distance along the relative track divided by relative speed gives time to closest approach.

Oxford groups several reusable problem types with relative motion. They reinforce the same discipline: draw the geometry, select the correct ground or air quantity, then calculate.

Implications of Geometry on the Triangle of Velocities

A wind exactly at right angles to track does not necessarily give zero head or tail component, because heading and track are separated by drift. Strictly, zero along-track component occurs when the wind is at right angles to the bisector between heading and track.

Calculation of Rhumb Line Track Angles

For short and moderate mid-latitude problems, form a right triangle from change of latitude and departure. From 45 N 010 W to 48 degrees 30 minutes N 015 W, change of latitude is 210 NM. Five degrees of longitude at about 45 N gives departure 300 x cos 45 degrees = 212.1 NM west. The angle north of west is arctangent 210 / 212.1 = 44.7 degrees, giving rhumb-line track about 315 T.

Cross-track displacement between Rhumb Line and Great Circle tracks

A rhumb line and great circle between the same endpoints separate most near the middle of the route. Use the applicable conversion-angle and 1-in-60 relationships only where the small-angle assumptions remain valid.

Timing to a Beacon

Timing depends on groundspeed, but the crew changes TAS, CAS or Mach. Therefore find current groundspeed and the wind component before changing airspeed. In Oxford's worked case, 150 NM in 17 minutes gives GS 530 kt. Mach 0.77 at minus 55 Celsius gives TAS about 442 kt, so tailwind is 88 kt. To cover 150 NM in 21 minutes needs GS 429 kt, hence TAS 341 kt and Mach about 0.59.

PNR is a fuel and endurance limit. Before using its formula, separate usable trip endurance from fuel that must remain protected for reserve, holding or diversion.

Endurance

If 3600 kg is available for the out-and-return decision and fuel flow is 600 kg per hour, safe endurance E is 6 hours. Fuel protected as reserve is not included in E.

Safe endurance

Safe endurance is the time available for the defined outbound and return operation after required reserves and unusable fuel have been excluded. If start, climb, descent or return flows differ, account for them separately rather than forcing one average into every segment.

Fuel rather than time

When outbound and homebound fuel flows differ, solve in fuel units. For a turn point at distance x, outbound fuel is x divided by outbound GS, multiplied by outbound fuel flow. Return fuel is x divided by homebound GS, multiplied by homebound fuel flow.

With 3000 kg available, outbound flow 500 kg per hour at 400 kt and homebound flow 400 kg per hour at 500 kt, distance = 3000 / (500 / 400 + 400 / 500) = about 1463 NM.

The point of no return, also called the point of safe return in planning questions, is the farthest point from which the aircraft can return to base within its safe endurance.

Point of No Return and Point of Safe Return

Let E be safe endurance in hours, O outbound groundspeed and H homebound groundspeed. If time outbound to the turn point is t, return time is distance divided by H, or tO / H. At the limiting point, outbound time plus return time equals E.

Worked PNR calculation

Safe endurance is 8 hours, outbound GS is 450 kt and homebound GS is 350 kt. Time to PNR = 8 x 350 / 800 = 3.5 hours. Distance from base = 3.5 x 450 = 1575 NM. The return takes 1575 / 350 = 4.5 hours, giving exactly 8 hours total.

Operational interpretation

Before PNR, a return within the stated safe endurance remains possible under the assumed conditions. Beyond it, the planned return is no longer possible with the protected reserve intact. Recalculate if wind, routing, fuel flow or available fuel changes.

Wind changes outbound and homebound groundspeeds in opposite senses. The PNR formula must use the two actual groundspeeds, not TAS twice.

Effect of wind on PNR time

With a headwind outbound, O decreases and H increases, so the aircraft takes longer to reach the PNR. With a tailwind outbound, O increases and H decreases, so PNR time is shorter.

Effect of wind on PNR distance

For equal TAS in both directions and a constant wind component, PNR distance contains the product O x H. Any non-zero along-track wind reduces that product compared with still air. Reversing the wind direction changes the time to PNR but gives the same PNR distance when only the sign of the component changes.

Worked wind comparison

Changing fuel flow

If homebound altitude or power changes fuel flow, use the fuel-distance equation rather than the constant-endurance shortcut. Include any fixed climb or descent fuel before allocating the remaining decision fuel.

The critical point, also called the equal-time point, is the position from which time to continue to one endpoint equals time to return to the other endpoint.

Critical Point and Equal-Time Point

Let D be total route distance, O groundspeed from base to destination and H groundspeed from destination towards base. If x is distance from base to CP, return time is x / H and onward time is (D minus x) / O. At CP these times are equal.

Worked CP calculation

The route is 2400 NM, outbound GS is 480 kt and homebound GS is 320 kt. CP distance from base = 2400 x 320 / 800 = 960 NM. Return time is 960 / 320 = 3 hours. Distance onward is 1440 NM and onward time is 1440 / 480 = 3 hours.

What CP does not use

CP depends on total route distance and the two groundspeeds. It does not require endurance or fuel. Fuel may decide whether either option is operationally possible, but it does not define the equal-time geometry.

PNR and CP respond differently to wind. PNR is limited by endurance; CP is positioned by equal travel time between the route endpoints.

Effect of wind on the critical point

The critical point moves towards the upwind end of the route. With a tailwind outbound, the base is upwind and CP moves towards base. With a headwind outbound, the destination is upwind and CP moves towards destination.

Zero-wind benchmarks

When O equals H, CP lies halfway along the route. PNR time is half the safe endurance and PNR distance is TAS multiplied by half the safe endurance.

A disciplined layout prevents the common mistakes: using TAS instead of groundspeed, putting the wrong speed in the numerator and confusing an endurance point with an equal-time point.

Worked route

Route distance D is 1200 NM. TAS is 300 kt, wind component is 60 kt tail outbound, and safe endurance E is 7.5 hours. Therefore O = 360 kt and H = 240 kt.

1. PNR time = 7.5 x 240 / 600 = 3.0 hours.
2. PNR distance = 3.0 x 360 = 1080 NM from base.
3. Return time from PNR = 1080 / 240 = 4.5 hours. Total time is 7.5 hours.
4. CP distance = 1200 x 240 / 600 = 480 NM from base.
5. Return time from CP = 480 / 240 = 2 hours.
6. Onward distance is 720 NM and onward time = 720 / 360 = 2 hours.

Reverse the wind

With 60 kt headwind outbound, O = 240 and H = 360. PNR time becomes 4.5 hours but PNR distance remains 1080 NM. CP moves to 720 NM from base, towards the now-upwind destination. Return and onward time from CP remain 2 hours.

Problem-solving checklist

Formula summary